Question

The product of any $r$ consecutive natural numbers is always divisible by

• A. $r!$
• B. $r^2$
• C. $r^n$
• D. none of these

Answer 1

The correct option is A

$r!$

Explanation for the correct option:

Product of $r$ consecutive terms$=n(n-1)(n-2)....(n-r+1)$

We know ${}^n{C}_r=naturalnumber$

$\begin{array}{rcl}{}^n{C}_r& =& \frac{n!}{r!(n-r)!}=\frac{n(n-1)(n-2)(n-3)....(n-r+1)...3.2.1}{r!(n-r)(n-r-1)(n-r-2).......3.2.1}\\ & =& \frac{n(n-1)(n-2)(n-3)....(n-r+1)(n-r)(n-r-1)(n-r-2)...3.2.1}{r!(n-r)(n-r-1)(n-r-2).......3.2.1}\\ & =& \frac{n(n-1)(n-2)(n-3)....(n-r+1)}{r!}\\ & =& \frac{Productofrconsecutiveinterger}{r!}=naturalnumber\end{array}$

Hence we can say product of any $r$ consecutive natural numbers is always divisible by $r!$

Therefore the product of $r$ consecutive natural numbers is divisible by the factorial of $r$

Hence, option A is correct